New answers tagged

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To calculate the minimum energy needed for the reaction the products are assumed to be stationary, i.e. the momentums are zero. With $ \pmb p^2 = E^2 - \vec p^2 = E^2 = m^2$ follows: $$({p_1^\mu}' + {p_2^\mu}' + {p_3^\mu}' + {\bar p_4^\mu}')^2 = (E_1 + E_2 + E_3 + E_4)^2 - (\vec p_1 + \vec p_2 + \vec p_3 + \vec p_4)^2 = (E_1 + E_2 + E_3 + E_4)^2 = (m_p + ...


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but still we have momentum telling us that both blocks must rise to the same height That's not true here. In the first case where the bullet is embedded, the final velocity of the bullet and the block must be identical. Since initial momentum of the two shots were the same, then the final momentum will be the same as well. Because they are connected, ...


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The simplest approach to a problem like this would assume that the collision is elastic, and that you have some knowledge of the elastic constant. But a collision between car and human is not that. Instead, let us assume that the "elbow sized object" hits the human in the mid section, and that it doesn't simply go right through him. Then the next thing that ...


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For inelastic collision, the ball deformation usually accompanies heat generation. To keep the system temperature constant, there should be some heat loss. This will increases the system's entropy.


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Very generally speaking, a crater is about 10 times the diameter of the meteor, with a direct hit. so at some 950 km in diameter, we can guesstimate a crater roughly covering 9,500 km, which is 1/4 the way around the Earth. If we give an impact speed of slightly greater than escape velocity of 12-13 km/s, it would take over a minute to complete it's ...


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In both cases, the speed of the body at time $t=0$ would be zero, if it was initially at rest. Then you will ask, what is the difference between an impulsive force and an "ordinary force" which keeps acting continuously? Actually, there is no fundamental difference between them. But when we say impulsive force, we mean that the force is of very large ...


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Considering momentum: $$m_1u_1+m_2u_2=m_1v_1+m_2v_2$$ Therefore, $$m_1u_1-m_1v_1=m_2v_2-m_2u_2$$ Factorising gives us: $$m_1(u_1-v_1)=m_2(v_2-u_2)$$ Allowing us to rearrange to: $$\frac{m_{1}}{m_{2}}=\frac{v_{2}-u_{2}}{u_{1}-v_{1}}$$ Using the fact that you say the velocities can be interchanged, we obtain a final answer of: ...


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We consider friction to an impulsive force, in cases when normal force is impulsive. Here's how:We know that $f=\mu N$(only during slipping motion, for no slipping frictional force is equal to applied force RESISTING friction). Since friction is proportional to normal reaction, it will be impulsive only when normal force is impulsive.Thus, if in a situation ...


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You can only have an inelastic collision between two bodies if one or both of the bodies have some internal degrees of freedom that can absorb energy. For example if you have a rigid sphere then the only type of energy it can possess is kinetic energy. If we collide two rigid spheres then conservation of energy means the sum of the kinetic energies before ...


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The coefficient of restitution tells you about the energy lost in the collision. Specifically e^2 is the ratio of the kinetic energy after to before the collision in the zero momentum frame. This depends not only on the elastic properties of the material, but also the structure of the body. If you take a very simple example and have 2 springs hit each ...


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The observation of granular convection as discussed here depends on the force of gravity. What you are seeing is the settling of the small granules according to the force that the sides of the bowl are imposing. Try hitting in one direction only. When shaken, the particles move in vibration-induced convection flow; individual particles move up through ...


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Yes, there are more unknowns than equations. You do not have sufficient information to solve for the requested quantities. Someone might be playing a prank on you! In reality, each ball and each paddle would have a specific finite stiffness, and one could use this information along with some clever math to determine the final velocities of all the bodies. ...


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As it happens, auto manufacturers (and the NHTSA I believe) have rules about maximum allowable jerk during cruise control, automatic braking, and similar machine-based events. (Trust me -- I work for an automotive active safety product company) Certainly airbags are required to have a maximum 'acceleration' as they open up, to reduce (and sadly, not ...


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For a 2D planar simulation with zero friction do the following Definitions Each body has 3 degrees of freedom. These are $(x_1,y_1,\theta_1)$ and $(x_2,y_2,\theta_2)$ defined at the center of mass. Each body has mass and mass moment of inertia. These are $m_1$, $m_2$ and $Iz_1$, $Iz_2$. The contact is at point A with coordinates $(x_A,y_A)$ and with ...


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No. Momentum is still conserved. In particular, the component of momentum parallel to the ground is conserved. So if the ball is going to the right before hitting the ground, it will continue going to the right after. The formula you refer to is for one-dimensional collisions. That applies only if the elements are arranged so that there actually is ...


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This cant be fully derived but a part of it can be. In an elastic collision kinetic energy is conserved, so 1/2*m1*u1^2 + 1/2*m2*u2^2 = 1/2*m1*v1^2 + 1/2*m2*v2^2 m1*u1^2 + *m2*u2^2 = *m1*v1^2 + *m2*v2^2 m1*u1^2 - *m2*v1^2 = *m2*v2^2 - *m2*u2^2 m1(u1+v1)(u1-v1)=m2(v2+u2)(v2-u2) ---------- equation 1 NOW, Accordinbg to conservation of linear ...


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If black holes can swallow any object, Black holes have a very strong gravitational field, this field from a distance attracts any other matter and the Newtonian approximation when there is a distance is adequate. When two strong gravitational fields get close enough General Relativity equations have to be used, but the "attraction" is there and very ...


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Work (or energy) is transferred from one particle to another, but the net effect is no change overall. How? Consider a collision force acting between two particles over a small time frame. During that time frame the particles move, and the work done on one particle is ${\rm d}W_1 = \boldsymbol{F} \cdot {\rm d}\boldsymbol{x}_1$. Since an equal and opposite ...


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The collision doesn't happen at a single point in space - rather the colliding objects exert a forces on each other over a distance as they approach and the recede. Consider a tennis ball hitting a racket - the ball and the strings of the racket deform and we get an increasing elastic restoring forces until the two objects at at their closest approach. ...


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Let I be the impulse transferred to the string due to the fall of the ball on the plank. Now as gravity is a non stop impulsive force so for calculation purpose it could be neglected. And the impulse would be transferred on the other side of the string as well. Let P' be the final momentum and p be the initial momentum. So,I=P'-P I=2mV-mu. -1 And, ...



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